A358235 - OEIS

A358235

Number of ways n' (the arithmetic derivative of n) can be formed as a sum (x * y') + (x' * y) from two factors x and y of n, with x <= y, so that the said sum does not involve any carries when the addition is done in the primorial base.

1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 2, 3, 2, 1, 1, 1, 1, 1, 4, 1, 3, 1, 2, 2, 2, 1, 3, 1, 3, 1, 1, 1, 1, 1, 2, 1, 3, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 1, 1, 3, 2, 1, 1, 1, 1, 2

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OFFSET

1,4

COMMENTS

Number of ways n can be factored as x*y, with 1 <= x <= y, so that the sum (x * y') + (x' * y) is carry-free when the addition is done in the primorial base. Here u' stands for A003415(u), the arithmetic derivative of u.

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for sequences related to primorial base

FORMULA

a(n) = Sum_{d|n} [d <= (n/d) and A329041(d*A003415(n/d), A003415(d)*(n/d)) == 1], where [ ] is the Iverson bracket, and the dyadic function A329041 returns 1 only when its two arguments do not generate any carries when added together in the primorial base.

For all n >= 1, a(n) <= A038548(n). [See A358673 for the indices where the equality is attained].

For all n >= 1, a(A100484(n)) = A358233(A100484(n)).

EXAMPLE

a(6) = 2 because 6 has only two factor pairs, {1, 6} and {2, 3}, and for both of those pairs the criteria is satisfied, as we have A329041(1*A003415(6), A003415(1)*6) = A329041(5, 0) = 1 and A329041(2*A003415(3), A003415(2)*3) = A329041(2, 3) = 1. In the latter case the primorial base expansions of 2 and 3 are "10" and "11" (see A049345), which can be added together cleanly (i.e., without carries) to obtain "21" = A049345(2+3).

a(24) = 3 because 24 can be factored into two factors in four possible ways: 1*24, 2*12, 3*8 and 4*6, of which all others, except pair {4,6} are carry-free: we have A003415(6)*4 = 20 and A003415(4)*6 = 24, with respective primorial base expansions "310" and "400", which when added together, yield a carry at the third digit position from the right, because 3+4 = 7 > 4 (which is the max. allowed digit in that place), and therefore a(24) = 4-1 = 3.

a(63) = 3 because 63 can be factored into two factors in three possible ways: 1*63, 3*21 and 7*9. The trivial factorization always yields a carryless sum, and in this case also the other two factorizations satisfy the criteria: A329041(3*A003415(21), A003415(3)*21) = A329041(30, 21) = 1 (as 21 is "311" and 30 is "1000" in primorial base), and A329041(7*A003415(9), A003415(7)*9) = A329041(42, 9) = 1 (as 9 is "111", 42 is "1200", and 9+42 = 51 is "1311" in primorial base).

PROG

(PARI)

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));

A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };

A327936(n) = { my(f = factor(n)); for(k=1, #f~, f[k, 2] = (f[k, 2]>=f[k, 1])); factorback(f); };

A329041sq(row, col) = A327936(A276086(row)*A276086(col));

A358235(n) = sumdiv(n, d, ((d <= (n/d)) && 1==A329041sq((d*A003415(n/d)), (A003415(d)*(n/d)))));

CROSSREFS

Cf. A003415, A049345, A100484, A276086, A329041, A358672, A358673, A358674.

Cf. also A358233.

Sequence in context: A300978 A378375 A156144 * A136044 A375736 A184240

Adjacent sequences: A358232 A358233 A358234 * A358236 A358237 A358238

KEYWORD

nonn,base

AUTHOR

Antti Karttunen, Nov 26 2022

STATUS

approved